Maneela, Raghu and Aravind have some jems each. Five times the number of jems with Raghu equals seven times the number of jems with Maneela, and five times the number with Maneela equals seven times the number with Aravind. What is the minimum total number of jems all three of them can have together?

Introduction / Context:
This question involves ratios and least common multiples to find the smallest integer solution that satisfies two linked proportional relationships. Each person holds some number of jems, and there are conditions relating multiples of the counts. We must find the minimum total number of jems that satisfies both ratio conditions exactly.

Given Data / Assumptions:
• Let Maneela have M jems, Raghu have R jems and Aravind have A jems.
• Condition 1: 5R = 7M.
• Condition 2: 5M = 7A.
• All numbers M, R and A must be positive integers.
• We seek the minimum possible value of M + R + A under these constraints.

Concept / Approach:
From 5R = 7M, we get R = (7/5)M, so M must be a multiple of 5 for R to be an integer. From 5M = 7A, we get A = (5/7)M, so M must also be a multiple of 7 for A to be an integer. Therefore, M must be a multiple of the least common multiple of 5 and 7, which is 35. Taking M = 35 gives the smallest positive integer solution that satisfies both relations. Then we compute R and A and sum them to get the minimum total.

Step-by-Step Solution:
Step 1: From 5R = 7M, write R = 7M / 5.Step 2: From 5M = 7A, write A = 5M / 7.Step 3: For R to be an integer, M must be divisible by 5; for A to be an integer, M must be divisible by 7. So M is a multiple of LCM(5, 7) = 35.Step 4: Take the smallest possible M = 35.Step 5: Then R = 7M / 5 = 7 * 35 / 5 = 7 * 7 = 49.Step 6: Also, A = 5M / 7 = 5 * 35 / 7 = 5 * 5 = 25.Step 7: Total jems = M + R + A = 35 + 49 + 25 = 109.

Verification / Alternative check:
Check both conditions: 5R = 5 * 49 = 245 and 7M = 7 * 35 = 245, so the first condition is satisfied. Also, 5M = 5 * 35 = 175 and 7A = 7 * 25 = 175, so the second condition is satisfied. Since we chose the smallest possible M that is a multiple of both 5 and 7, the corresponding totals are minimal.

Why Other Options Are Wrong:
Options 108, 107 and 110 are close to 109 but cannot be expressed as M + R + A with positive integers satisfying both 5R = 7M and 5M = 7A. Any smaller M than 35 would make at least one of R or A fractional, which is not allowed. Therefore, 109 is the only feasible minimum total.

Common Pitfalls:
Students sometimes try random values and check conditions, which is inefficient and error prone. Others mistakenly think they can take M as 5 or 7 without checking integrality for both R and A. The correct approach is to use the least common multiple to ensure that all ratios yield integers. Carefully using LCM and verifying both conditions avoids these mistakes.

Final Answer:
The minimum total number of jems that all three can have together is 109.